Because there are two hydrogen atoms in every H2O, there are 3.0 moles of hydrogen in the water which was created, and thus, 3.00 moles of H in the original compound.
The original compound must have a C:H ratio of 1:3.
gravimetric analysis is a common way to use precipitation reactions in order to make quantitative determinations about the identity of an unknown sample.
A sample of an unknown compound is mixed with a solution that will cause one of the strontiums from the unknown sample to come out.
If the mass of the ion in the unknown is known, a mass percent calculation can be made.
It's easy to demonstrate due to the large amount of math involved.
A 4.33 g sample of a compound is dissolved in water.
You need to convert the moles of hydroxide into grams.
If 1.33 g of OH- is present in the sample, then there are also 1.33 g of OH- in the original sample.
To determine the identity of the sample, you have to know the mass percent of hydroxide in all of the possible compounds.
The mass of one mole of hydroxide is divided by the mass of the compound.
The most likely identity of the original unknown is the hydroxide mass percent in KOH.
The enthalpy is a measure of the energy that is released or absorbed by a substance when bonds are broken.
There is a negative enthalpy change.
Endothermic reactions have positive enthalpy changes.
All substances are stable in their lowest energy state.
Exothermic processes are more likely to be favorable than endothermic processes.
The diagram shows the energy change during an exothermic reaction.
The graph shows the amount of energy the reactants start with.
The highest point on the graph is where the reactants must have enough energy to reach the transition state.
At this point, all reactant bonds have been broken, but no product bonds have been formed, so this is the point in the reaction with the highest energy and lowest stability.
We reach the energy level of the products when we move to the right past the activated complex.
The reverse reaction for an exothermic reaction can be read in both directions.
A catalyst speeds up a reaction by giving the reactants an alternate pathway that has a lower activation energy, as shown in the diagram above.
The energy of the activated complex is lower for the catalyzed reaction than it is for the uncatalyzed reaction.
It has no effect on the equilibrium conditions if a catalyst lowers the activation energy for both the forward and reverse reaction.
The electrons are transferred between the reactants.
There are several rules for assigning oxidation states.
The oxidation state of any neutral atom is zero.
An oxidation state is equal to the charge on the ion.
Any kind of ionic compound includes ion bonding.
Iron is +3 and chlorine is -1.
Oxygen is -2.
Oxygen is -1 in hydrogen peroxide.
Oxygen is -2.
H2O is +2 when bonded to a nonmetal.
In CH4 hydrogen is +2 when it is bonded to a metal.
H2 is -1 in NaH.
The most common charge in a compound will oxidize in the absence of oxygen.
There are examples of fluorine being -1.
Sulfur is -2.
The oxidation states on elements in a neutral compound must add up to zero.
The oxidation states on elements in a polyatomic ion must add up to the charge on that ion.
Rule 6 is only used to determine the oxidation state of an element which does not fall under the first five rules.
Not many nonmetals with low electronegativity values have a lot of oxidation states.
Every element in H2SO4 has an oxidation state.
The total charge on the compound must be 0.
S will be found if you solve for it.
Determine the oxidation state of every element.
The oxidation states of some of the reactants are changed over the course of the reaction.
The oxidation state of Fe changes from 0 to +2.
H was reduced.
Fe was oxidized.
Reduction and oxidation go hand in hand.
One atom must be gaining electrons.
There is a device that might be useful.
A titration involves adding a solution at a known concentration to another solution of unknown concentration in order to determine the concentration of the unknown solution.
A color change is used to determine the end of a titration.
Titrations are often used in acidbase reactions, but they can also be used in redox reactions.
There is a lot of use of Potassium permanganate in redox titrations.
The permanganate ion has an oxidation state of +7 and the solution is a deep purple color.
When mixed with compounds that can oxidize, the oxidation state of the manganese in MnO - 4 is changed to +2.
It is a ion.
The end of the titration is marked by the solution turning pink.
The permanganate ion takes electrons from the compound and reduces them to Mn2+.
The extra permanganate ion that is added remains unreduced and retains its purple color once the compound that is being oxidation runs out.
A solution of Na2C2O4 is poured into an Erlenmeyer flask.
The solution in the flask is originally white, but turns pink after 14.86 liters of permanganate is added.
The first thing to do is balance the half-reactions.
The reduction half-reaction must be multiplied by 2 in order to balance the electrons with those in the oxidation half-reaction.
Five moles of oxalate are required for every mole of permanganate that reacts.
You won't be expected to memorize the various color changes that can occur during a redox titration.
If you are provided with the colors of the various compounds that are present during a titration, you can use a color change to determine an endpoint.
Every half-reaction has an electric potential.
The values for the standard reduction potential of half-reactions will be given to you.
You can read potentials in reverse and get oxidation potentials if you flip the sign on the voltage.
If you read the reduction half-reaction in reverse, you can get the oxidation potential for Zn.
The bigger the potential for a half-reaction, the more likely it is to happen.
If you know the potentials for the two half-reactions, you can calculate the potential of a redox reaction.
There are two things to remember when calculating the potential of a reaction.
Add the potential for the oxidation half-reaction to the reduction half-reaction.
The potential for a half-reaction by a coefficient is never calculated.
We ignored the fact that silver has a coefficient of 2 in the balanced equation.
The reduction strengths of two different metals can be determined qualitatively.
When a solid metal is placed into a metallic solution and a new solid starts to form, the reduction potential of the metal in solution is greater than that of the solid metal.
The reduction potential of the metal is higher if there are no solid forms.
Platinum is used as a conductor in an electrolytic cell because it is unreactive with most solutions.
In a galvanic cell, the oxidation and reduction reactions take place in separate chambers, and the electrons that are released by the oxidation reaction pass through a wire to the chamber where they are consumed.
There's a way to remember that.
The electrical neutrality in the cell is maintained by the salt bridge.
Positive cations from the salt bridge solution flow into the half-cell when the solution becomes less positively charged.
Negative anions from the salt bridge solution flow into the half-cell when oxidation occurs and the solution becomes more positively charged.
The solutions in the halfcells would become unbalanced if the salt bridge were to be removed.
The bridge is filled with water.
The Na+ cations and the Cl- anions flow as the cell functions.
To determine what would happen to the cell potential under nonstandard conditions, you need to learn more about equilibrium, which will be covered in Chapter 8.
An unfavored redox reaction can take place when an outside source of voltage is used.
The ion or the water molecule can be reduced or oxidized when a chemical is dissolved in water.
The nickel solution is an example.
If you compare the reduction potential of the cation with that of water, you can determine which substance is reduced.
The half-reaction with the more positive value will occur.
The anion's oxidation potential must be examined.
The half-reaction with the most positive value will occur.
When you flip the equation to an oxidation, remember that potentials are always given as reduction potentials.
The metal bars that conduct current and do not take part in the reaction are the anode and cathode.
Solid nickel is being created at the cathode and oxygen gas is being evolved at the anode.
A molten compound or pure water can be used to run a current.
You don't need to determine which reactions are taking place, as you only have one choice.
The galvanic cell is the same as the electrolytic cell in that it is subject to the An Ox/Red Cat rule.
Thelytic cells are used.
There are four steps to figuring out electrolysis problems.
The charge can be calculated in coulombs if you know the current and time.
You know how many electrons were involved in the reaction when you know the charge in coulombs.
3 is the number of electrons in the mole.
If you know the number of moles of electrons and half-reaction for the metal, you can figure out how many moles of metal are plated out.
If you know the number of moles of metal, you can use that information to calculate the number of grams of metal.
There are many different conversions before you come up with your final answer, so it's important to keep track of your units.
If all of your conversions are set up correctly, your final answer will have the correct units.
The following rules can be used to answer questions.
Salts with carbonate anions are insoluble.
No precipitate would form.
The correct net ionic equation is used to represent the reaction that occurs when solutions of potassium carbonate and copper are mixed.
A strip of metal X is placed into a solution and there is no reaction.
Metal Z starts to form on the strip when metal X is placed in a separate solution.
The activation energy is always more than the exothermic reaction.
The rate of the reaction will be increased by a catalyst.
It is above the energy level of the reactants, but below the energy level of the products.
It is below the energy level of the reactants, but above the energy level of the products.
The products and reactants have the same energy level.
The products and reactants have different levels of energy.
There are excess PO4 cations after the reaction, and the K3PO4 contains a cation that cannot form a precipitate.
The following information can be used to answer questions.
The molecule will lose more energy when they collide.
It will increase by a factor of four.
The bond energy in the reactants is reduced by a catalyst.
The energy differential between the reactants and products is reduced by a catalyst.
A catalyst lowers the activation energy of the reaction.
A sample of CuSO4 with a mass of 250 grams was heated until all the water was removed.
The sample had a mass of 160 grams.
The pressure was measured at 2 atm and the mixture contained 1 mole of CH4 and 2 moles of O2.
The gases underwent a reaction.
The oxidation number is the same.
Over a period of 600.0 seconds, a constant current of 5.00 amperes is applied to molten AlCl3.
Pennies are made mostly of zinc and coated with a thin layer of copper through electroplating.
The solution in the beaker is a strong acid and the cell is wired so that the zinc penny is the cathode.
The following reduction potentials can be used to answer questions.
2.54 g of beryllium chloride is dissolved into 50.00 mL of water inside a beaker.
You should include at least one beryllium ion, one chloride ion, and four water molecule in your diagram.
Make sure the atoms and ion are oriented in the same way.
H2O2 is a common disinfectant.
Pure hydrogen peroxide is a very strong oxidizer and isDiluted with water to low percentages before being bottled and sold.
If you want to know the concentration of H2O2 in a bottle of hydrogen peroxide, you can use acidified potassium permanganate.
During a titration, a student measures out a small amount of hydrogen peroxide solution into a graduated cylinder and then pours it into a flask.
Acetylene is a fuel used in welding.
The process of burning acetylene is exothermic.
A student performs an experiment in which a bar of unknown metal is placed in a solution with a formula.
The solution of CuSO4 is used to connect the metal to the copper bar.
The cell is linked together by a salt bridge.
Solid copper starts to form on the bar when a bar of metal is placed in the copper sulfate solution.
There is no visible reaction when a bar of copper is placed in the MNO3 solution.
Write the net ionic equation in the Cu/M cell.
Justify your answer.
Any ionic movement occurring in the salt bridge is indicated.
Justify your answer.
A current of 2.20 A is run through the two electrodes that were inserted into the solution.
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